| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 7 May 2007 13:07:47 IST
|
|
|
A wire of length l and mass m is bent in the form of a recangle ABCD with AB/CD=2. The moment of inertia of this wire frame about the side BC is? plz answer soon friends..
|
|
|
|
7 May 2007 13:42:52 IST
|
|
|
are u sure AB/CD=2??
If it is so it can't be a rectangle!!
|
Impossible is Nothing |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
7 May 2007 13:58:21 IST
|
|
|
ys right i think he has done some thyping mistake it shouid be AB/BC=2
if it is then
I= ml2/27
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
7 May 2007 14:16:02 IST
|
|
|
Is answer 5ML2/144
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
7 May 2007 15:33:19 IST
|
|
|
well i got 7ML2/162 if AB:BC=2.. whats the correct ans ranveer?
|
Impossible is Nothing |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
7 May 2007 15:36:15 IST
|
|
|
yup even i got 7ml^2/162
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
7 May 2007 17:17:27 IST
|
|
|
Hey how did u get it?
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
7 May 2007 17:52:04 IST
|
|
|
the answer shouid be 5ml^2/432 is u r taking axis along the longer side but ml^2/27 if u r taking axis along shorter side
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
7 May 2007 18:13:02 IST
|
|
|
here how
the mass of side AB=mass of side CD=m/3
mass of side BC=mass of side AD=m/6
lenght of side AB=L/3=lenght of side CD
lenght of side BC=lenght of side AD=L/6
distance of centre of mass of side AB=distance of centre of mass of side from axis of rotation = L/6 bcoz centre of mass will be at the centre of side
distance of centre of mass of side BC=distance of centre of mass of side AD=L/3
now moment of inertia due to side AB= moment of inertia due to side CD=M*(L/6)^2 M=m/3 so it will come (m/3)*L^2/36=mL^2/108
moment of inertia due to both side =2*mL^2/108=mL^2/54
NOW moment of inertia due to side AD=M" * (L/3)^2 =(m/6)^2 *L^2 /9=mL^2/54 here M"=m/6 as already mention
NOTE- moment of inertia due to side BC=0
TOTAl inertia=2*(ML^2/54)=mL^2 /27
similarly when we will take axis along longer side it will come 7mL^2/162
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
7 May 2007 20:58:19 IST
|
|
|
sorry ... ques was AB/BC=2 ..... nd the correct answer is 7ml^2/162 according to the book
rphy & pkg r correct ... friends plz help me by posting the procedure u followed
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
ok now the mass of AB=mass of DC = m/3
mass of AD=mass of BC= m/6
length of AB = length of CD = L/3
length of AD = length of BC = L/6
MOI of AB about BC = MOI of CD about BC = (m/3)*(L/3)2*(1/3)
MOI of AD about BC = (m/6)*(L/3)2 //(Use MOI=mr2 as AD is at a constant distance from BC)
Adding the 3, final MOI = 2*(m/3)*(L/3)2*(1/3) + (m/6)*(L/3)2
= 7mL2/162
|
Impossible is Nothing |
this reply: 7 points
(with 1
in 2 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
